What number is 160% of 75

Treble 29 = 160- ? By the way, treble is triple. Please answer correctly! This is for Maths.

nika2105 [10]

29 = 160 - x

x = ?

160 - 29 = 131

SO x = 131

Check:-

160 - 131 = 29

SO,

29 = 160 - 131.

Hope i helped ya!

7 0

2 years ago

H(n) = -2n² + 4; Find h(4) A.-28 B.68 C.28 D.-4

Elena L [17]

Answer:

-28

Step-by-step explanation:

h(4)=-2(4)^2+4

-32+4=-28

7 0

3 years ago

6.

larisa86 [58]

A. Every month Population will increase by a factor of 0.84%.

B. Every 3 months Population will increase by a factor of 2.5%.

C. Increase in population in every 20 months is 10% + 6.72% = 16.72%.

Step-by-step explanation:

Here, we have number of employees in a company has been growing exponentially by 10% each year. So , If we have population as x in year 2019 , an increase of 10% in population in 2020 as $x + \frac{x}{100}\times10$ which is equivalent to $\frac{11x}{10}$ .

A.

For each month: We have 12 months in a year and so, distributing 10% in 12 months would be like $\frac{10}{12} = 0.84$ . ∴ Every month Population will increase by a factor of 0.84%.

B.

In every 3 months: We have , 12 months in a year , in order to check for every 3 months $\frac{12}{3} = 4$ and Now, Population increase in every 3 months is $\frac{10}{4} = 2.5$ . ∴ Every 3 months Population will increase by a factor of 2.5%.

C.

In every 20 months: We have , 12 months in a year in which increase in population is 10% . Left number of moths for which we have to calculate factor of increase in population is 20-12 = 8. For 1 month , there is 0.84% increase in population ∴ For 8 months , 8 × 0.84 = 6.72 %.

So , increase in population in every 20 months is 10% + 6.72% = 16.72%.

5 0

3 years ago

HELP if you’re smart answer all 9 questions its easy NO LINKS!!! pls be correct

Nesterboy [21]

Step-by-step explanation:

if it's easy why don't you do it. not trying to be rude are whatever

8 0

2 years ago

A researcher wishes to see if the five ways (drinking decaffeinated beverages, taking a nap, going for a walk, eating a sugary s

e-lub [12.9K]

Answer:

$\chi^2 = \frac{(21-12)^2}{12}+\frac{(16-12)^2}{12}+\frac{(10-12)^2}{12}+\frac{(8-12)^2}{12}+\frac{(5-12)^2}{12}=13.833$

$p_v = P(\chi^2_{4} >13.833)=0.00785$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(13.833,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the drowsiness are NOT equally distributed among office workers .

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Method Beverage Nap Walk Snack Other

Number 21 16 10 8 5

The total on this case is 60

We need to conduct a chi square test in order to check the following hypothesis:

H0: Drowsiness are equally distributed among office workers

H1: Drowsiness IS NOT equally distributed among office workers

The level of significance assumed for this case is $\alpha=0.1$

The statistic to check the hypothesis is given by:

$\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total}{5}$

And the calculations are given by:

$E_{Beverage} =\frac{60}{5}=12$

$E_{Nap} =\frac{60}{5}=12$

$E_{Walk} =\frac{60}{5}=12$

$E_{Snack} =\frac{60}{5}=12$

$E_{Other} =\frac{60}{5}=12$

And the expected values are given by:

Method Beverage Nap Walk Snack Other

Number 12 12 12 12 12

And now we can calculate the statistic:

$\chi^2 = \frac{(21-12)^2}{12}+\frac{(16-12)^2}{12}+\frac{(10-12)^2}{12}+\frac{(8-12)^2}{12}+\frac{(5-12)^2}{12}=13.833$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(catgories-1)=(5-1)=4$

And we can calculate the p value given by:

$p_v = P(\chi^2_{4} >13.833)=0.00785$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(13.833,4,TRUE)"

Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that the drowsiness are NOT equally distributed among office workers .

7 0

3 years ago